Question

What is the Laplace transform of \[t\cos at + \sin at\]?

Answer 1

Hint:To find the Laplace transform of \[t\cos at + \sin at\], we will use the one of the properties of Laplace transformation i.e., . Using this we will find the Laplace transformation of \[\sin at\] and then using this we will find the Laplace transform of \[t\cos at\] and then we will add these two results to find the Laplace transformation of \[t\cos at + \sin at\].

Complete step by step answer:
We have to find the Laplace transform of \[t\cos at + \sin at\]. For this we will use the rule for the Laplace transform of the derivative.
Here, \[L\left[ {f(t)} \right] = F(s)\]
Let \[f(t) = \sin at - - - (1)\]
Putting \[t = 0\], we get
\[ \Rightarrow f(0) = \sin \left( 0 \right)\]
On simplification we get,
\[ \Rightarrow f(0) = 0\]
On differentiating \[f(t) = \sin at\], we get
\[ \Rightarrow {f^1}(t) = a\cos at\]
Putting \[t = 0\], we get
\[ \Rightarrow {f^1}(0) = a\cos \left( 0 \right)\]
On simplification we get,
\[ \Rightarrow {f^1}(0) = a\]
Now, on differentiating \[{f^1}(t) = a\cos at\], we get
\[ \Rightarrow {f^{''}}(t) = - {a^2}\sin at - - - (2)\]
Using \[(1)\] in \[(2)\], we get
\[ \Rightarrow {f^{''}}(t) = - {a^2}f(t)\]
Taking the Laplace transformation of both the sides we get
\[ \Rightarrow L\left[ {{f^{''}}(t)} \right] = L\left[ { - {a^2}f(t)} \right] - - - (3)\]
We know $L\left[ {f'\left( t \right)} \right] = {s^2}F\left( s \right) - sf\left( 0 \right) - f'\left( 0 \right)$ . Using this we can write \[(3)\] as
\[ \Rightarrow {s^2}F(s) - sf(0) - {f^1}(0) = L\left[ { - {a^2}f(t)} \right] - - - (4)\]
As \[L\left[ {k \times f(t)} \right] = kL\left[ {f(t)} \right]\], where \[k\] is a constant.
Therefore, we can write \[(4)\] as
\[ \Rightarrow {s^2}F(s) - sf(0) - {f^1}(0) = - {a^2}L\left[ {f(t)} \right]\]
Putting the value of \[f(0)\], \[{f^1}(0)\] and \[L\left[ {f(t)} \right] = F(s)\], we get
\[ \Rightarrow {s^2}F(s) - 0 - a = - {a^2}F(s)\]
On rearranging we get
\[ \Rightarrow {s^2}F(s) + {a^2}F(s) = a\]
Taking \[F(s)\] common from the left hand side of the above equation, we get
\[ \Rightarrow \left( {{s^2} + {a^2}} \right)F(s) = a\]
Dividing both the sides by \[\left( {{s^2} + {a^2}} \right)\], we get
\[ \Rightarrow F(s) = \dfrac{a}{{\left( {{s^2} + {a^2}} \right)}}\]
\[\therefore L\left[ {\sin at} \right] = \dfrac{a}{{\left( {{s^2} + {a^2}} \right)}} - - - (5)\]
Similarly, now suppose that we let \[f(t) = t\cos at - - - (6)\]
Putting \[t = 0\], we get
\[f(0) = \left( 0 \right) \times \cos \left( 0 \right)\]
On simplification we get
\[ \Rightarrow f(0) = 0\]
On differentiating \[f(t) = t\cos at\] using the product rule of differentiation, we get
\[ \Rightarrow {f^1}(t) = \left( t \right)\left( { - a\sin at} \right) + \left( 1 \right)\left( {\cos at} \right)\]
On simplification we get
\[ \Rightarrow {f^1}(t) = - at\sin at + \cos at\]
Again, on differentiating we get
\[ \Rightarrow {f^{''}}(t) = \left( { - at} \right)\left( {a\cos at} \right) + \left( { - a} \right)\left( {\sin at} \right) - a\sin at\]
On simplification we get
\[ \Rightarrow {f^{''}}(t) = - {a^2}t\cos at - 2a\sin at\]
Using \[(6)\], we can write
\[ \Rightarrow {f^{''}}(t) = - {a^2}f(t) - 2a\sin at - - - (7)\]
Putting \[t = 0\], we get
\[ \Rightarrow {f^{''}}(0) = - {a^2}f(0) - 2a\sin \left( 0 \right)\]
On simplification we get
\[ \Rightarrow {f^{''}}(0) = 0\]
On taking Laplace transformation of \[(7)\], we get
\[ \Rightarrow L\left[ {{f^{''}}(t)} \right] = L\left[ { - {a^2}f(t) - 2a\sin at} \right]\]
As Laplace transform satisfy the linear property, so we can write the above equation as
\[ \Rightarrow L\left[ {{f^{''}}(t)} \right] = - {a^2}L\left[ {f(t)} \right] - 2aL\left[ {\sin at} \right]\]
Above we have proved that \[L\left[ {\sin at} \right] = \dfrac{a}{{\left( {{s^2} + {a^2}} \right)}}\]. Using this and putting the value of \[L\left[ {{f^{''}}(t)} \right]\] and \[L\left[ {f(t)} \right]\], we get
\[ \Rightarrow {s^2}F(s) - sf(0) - {f^1}(0) = - {a^2}F(s) - 2a \times \dfrac{a}{{{s^2} + {a^2}}}\]
Putting the value of \[f(0)\] and \[{f^1}(0)\], we get
\[ \Rightarrow {s^2}F(s) - 0 - 1 = - {a^2}F(s) - \dfrac{{2{a^2}}}{{{s^2} + {a^2}}}\]
On simplification, we get
\[ \Rightarrow {s^2}F(s) - 1 = - {a^2}F(s) - \dfrac{{2{a^2}}}{{{s^2} + {a^2}}}\]
On rearranging we get
\[ \Rightarrow {s^2}F(s) + {a^2}F(s) = 1 - \dfrac{{2{a^2}}}{{{s^2} + {a^2}}}\]
On simplification and taking common, we get
\[ \Rightarrow F(s)\left( {{s^2} + {a^2}} \right) = \dfrac{{{s^2} + {a^2} - 2{a^2}}}{{{s^2} + {a^2}}}\]
On further simplification we get
\[ \Rightarrow F(s)\left( {{s^2} + {a^2}} \right) = \dfrac{{{s^2} - {a^2}}}{{{s^2} + {a^2}}}\]
Dividing both the sides by \[\left( {{s^2} + {a^2}} \right)\] we get
\[ \Rightarrow F(s) = \dfrac{{{s^2} - {a^2}}}{{{{\left( {{s^2} + {a^2}} \right)}^2}}}\]
\[\therefore L\left[ {t\cos at} \right] = \dfrac{{{s^2} - {a^2}}}{{{{\left( {{s^2} + {a^2}} \right)}^2}}} - - - (8)\]
Adding \[(5)\] and \[(8)\], we get
\[ \Rightarrow L\left[ {\sin at} \right] + L\left[ {t\cos at} \right] = \dfrac{a}{{\left( {{s^2} + {a^2}} \right)}} + \dfrac{{{s^2} - {a^2}}}{{{{\left( {{s^2} + {a^2}} \right)}^2}}}\]
On simplifying we get
\[ \Rightarrow L\left[ {\sin at} \right] + L\left[ {t\cos at} \right] = \dfrac{{a\left( {{s^2} + {a^2}} \right) + \left( {{s^2} - {a^2}} \right)}}{{{{\left( {{s^2} + {a^2}} \right)}^2}}}\]
On further simplification, we get
\[ \Rightarrow L\left[ {\sin at} \right] + L\left[ {t\cos at} \right] = \dfrac{{a{s^2} + {s^2} + {a^3} - {a^2}}}{{{{\left( {{s^2} + {a^2}} \right)}^2}}}\]
On taking common,
\[ \Rightarrow L\left[ {\sin at} \right] + L\left[ {t\cos at} \right] = \dfrac{{{s^2}\left( {a + 1} \right) + {a^2}\left( {a - 1} \right)}}{{{{\left( {{s^2} + {a^2}} \right)}^2}}}\]
On rewriting we get
\[ \Rightarrow L\left[ {t\cos at + \sin at} \right] = \dfrac{{{s^2}\left( {a + 1} \right) + {a^2}\left( {a - 1} \right)}}{{{{\left( {{s^2} + {a^2}} \right)}^2}}}\].

Therefore, the Laplace transform of \[t\cos at + \sin at\] is \[\dfrac{{{s^2}\left( {a + 1} \right) + {a^2}\left( {a - 1} \right)}}{{{{\left( {{s^2} + {a^2}} \right)}^2}}}\].

Note:Laplace transform is an integral transform that converts a function of a real variable to a function of a complex variable. Laplace transform satisfy the linear property i.e., \[L\left[ {f\left( t \right) + g\left( t \right)} \right] = L\left[ {f\left( t \right)} \right] + L\left[ {g\left( t \right)} \right]\].